Advanced methods and approaches for solving Sudoku puzzles
Ok, so you mean that there by no means can be more than 2 groups of digits in a row, col or box and that there never is just one group (like in this example: 123,234,345,456,567,678,789,891,912)

sorry, this is what that m thing says...
Last edited by Bjorn on Tue Jun 07, 2005 2:20 pm, edited 1 time in total.
Bjorn

Posts: 9
Joined: 07 June 2005

On the contrary. In half-finished puzzles there is usually just 'one group' in each column, row and 3x3 block. They don't tell you anything - you have to look for those where there are two or more groups, in order for the observation to be worth anything to the puzzle (i.e. to reduce the possibilities in the row/column/block we are considering).

There can also be three groups (mathematically, there could be four pairs or three pairs and a triplet, but I've never seen that in a puzzle). Particularly if one or two of the groups are pairs, three is definitely possible. What I'm not sure about is whether, when there are three groups, how many of n type and how many of m type are usually present, as per Animator's definitions. Is there always at least one of each? It has been a while since I've seen a puzzle in with three groups to be honest.

The point Animator was making is that SuDoku row/column/boxes have a length of nine so if there are two groups, the biggest size of the smallest one is four. You therefore never need to look for groups that are bigger than four as if two groups exist, one will be no bigger than four.
abailes

Posts: 27
Joined: 02 June 2005

Please, can someone give me a hint how to proceed? What im I missing?

--------------------
=14*=6**=98*=
=***=19*=***=
=**7=***=5**=
--------------------
=**9=***=*62=
=62*=7*4=*59=
=58*=9**=***=
--------------------
=**2=***=3**=
=***=*79=***=
=*5*=**1=*94=
--------------------
Bjorn

Posts: 9
Joined: 07 June 2005

The hint you are looking for:
* first look at the number 2, where can it go on row 1?
* what does this mean for that particular box (the one that needs the 2 on row 1)?
* take a look at the bottom row of that box... do you see the triplet?
* take a look at column 8, write down the list of candidates
* remember what you said about the bottom row of the previous box, it is important.
* remove the appropriate candidates from column 8 (if you haven't do so yet)
* do you see the pair in column 8?
* fill in r7c8.
Animator

Posts: 469
Joined: 08 April 2005

Bjorn wrote:Ok, so you mean that there by no means can be more than 2 groups of digits in a row, col or box never is just one group (like in this example: 123,234,345,456,567,678,789,891,912)

(Please be careful with editing posts... as in, you need to make sure the following replies still make sense...)

As abailes said, the point I was trying to make is that there if there are only two groups (which is not a requirement) that the biggest group then has (maximum) 4 numbers.

Yes, there can be groups with more numbers, yes there can be a group of 9 numbers (as in your example), but it doesn't help you... so I really was talking about intresting groups.

To continue, yes there can be a group of 8 numbers, but the consecquence of that is that there is a digit on the row (column, ...) that occures only once. So you can see that as a group of 1 number... (or you fill it in without thinking about it).
Animator

Posts: 469
Joined: 08 April 2005

Animator,

Well done. You beat me to it (again).

I thought I had it, I thought there was a swordfish on the 4s, but when I went to check the logic there was a mistake. Just to put me off further, it gave me the right answer in the end (assuming that the answer is unique).

So close - next time...

I'm disappointed on two levels - it was the first time I've ever had to use swordfish in a puzzle and it turns out to be false!

Andrew
abailes

Posts: 27
Joined: 02 June 2005

Andrew,

You will never need a swordfish to solve a 'very hard' puzzles :(

I've seen 'very hard' puzzles which has a swordfish, but none of them were actualy useful.

If you really want some puzzles that requires the Swordfish technique, then you can look at the Sudoku Programmers forum: http://www.setbb.com/phpbb/viewtopic.php?t=18&mforum=sudoku (direct link to the Swordfish puzzles)
Animator

Posts: 469
Joined: 08 April 2005

Animator wrote:you never need a subset of 6 numbers/cells.

the maximum you'll ever need is 4 numbers/cells.

Why? The answer is simple, there are always two groups.

Either n cells share the same numbers (and only these numbers),

or there are m cells that are the only valid candidates for m numbers.

A little example of that: in column 9 of this puzzles there are two groups, one of 2 numbers, and one of 4 numbers.

The 'n'-example: the group: 1, 3, 6, 7. 4 numbers, 4 cells. All of these four cells have share (only) these numbers.

The 'm' example: the group 5, 8. There are only two cells that can have these as candidate. Ofcourse these cells has extra candidates aswell, but they are just irrelevant.

And I thought I was being subtle by using letters rather than numbers (well, at least for my initial explanation of the n=4 example) ;)

Animator - can I just say at this point that I really appreciate your input into these forums. I know I'm a relative newbie here, but the patience and understanding you show when helping others out is a great example for us all to follow. It's genuinely appreciated.

(but note one thing - if we both decide to enter the Times competition later this year, I intend to humiliate you )
whohe

Posts: 32
Joined: 28 May 2005

I could not agree more with whohe.
Bjorn

Posts: 9
Joined: 07 June 2005

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